Question

Suppose a snack bar is burned in a calorimeter and heats 2,000 g water by 20 °C. How much heat energy was released? (Hint: Use the specific heat equation.) Give your answer in both joules and calorie

Answer 1

Answer:

Explanation:

Remember that the key is the water so

using the formula q=m*s*deltT

We have the mass 2000g

we have the specific heat of water which is 1.00 cal/ g C

and the final temperature 20 C

Remember that delt T is the change in temperature meaning (Tfinal - Tinitial) Assuming that the water is at room  temperature (25 C)  before starting the reaction then

q= (2000 g)(1.00 cal/g C)(20-25)

q= -10, 000 cal  

if you want you can convert to joules by multiplying by 4.314 J

In more simpler terms - If all the heat released was absorbed by the water, and it takes 1 calorie of heat to raise

the temperature of 1 gram of water by 1°C that is a total of 20 times 2000 or 40,000 cal.

or 40 kilocalories. If you want it in joules, there are 4.184 joules to one calorie so it is

4.184 times 40 or 167.36 kilojoules.

Hope that helped

Answer 2

A snack that, when burned, heats 2,000 g of water by 20 °C, releases 4.0 × 10⁴ cal (1.7 × 10⁵ J).

A snack bar is burned in a calorimeter and heats 2,000 g  (m) of water by 20 °C (ΔT).

We can calculate the heat absorbed by the water (Qw) using the following expression.

$Qw = c \times m \times \Delta T = \frac{1cal}{g.\° C } \times 2,000 g \times 20 \° C = 4.0 \times 10^{4} cal$

where,

• c: specific heat capacity of water

According to the law of conservation of energy, the sum of the heat absorbed by the water (Qw) and the heat released by the combustion (Qc) is zero.

$Qw + Qc = 0\\\\Qc = -Qw = -4.0 \times 10^{4} cal$

Finally, we can convert -4.0 × 10⁴ cal to J using the conversion factor 1 cal = 4.18 J.

$-4.0 \times 10^{4} cal \times \frac{4.18 J}{1cal} = 1.7 \times 10^{5} J$

A snack that, when burned, heats 2,000 g of water by 20 °C, releases 4.0 × 10⁴ cal (1.7 × 10⁵ J).

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